Question

A spring is hung from the ceiling. When a block is attached to its end, it stretches 2.6 cm before reaching its new equilibrium length. The block is then pulled down slightly and released. What is the frequency of oscillation?

Answer 1

Answer:

f = 3.09 Hz

Explanation:

This is a simple harmonic motion exercise where the angular velocity is

         w² = $\frac{k}{m}$

to find the constant (k) of the spring, we use Hooke's law with the initial data

         F = - kx

where the force is the weight of the body that is hanging

        F = W = m g

we substitute

        m g = - k x

        k = $- \frac{m g}{x}$

we calculate

        k = $- \frac{9.8 m}{- 2.6 \ 10^{-2}}$

        k = 3.769 10² m

we substitute in the first equation

       w² = $\frac{ 3.769 \ 10^2 \ m }{m}$

       w = 19.415 rad / s

angular velocity and frequency are related

       w = 2πf

        f = $\frac{w}{2\pi }$

        f = 19.415 / 2pi

        f = 3.09 Hz